关于Laplace-Stieltjes变换所定义的函数增长性, 余家荣首先在文[1]中得到Valiron-Knopp-Bohr公式, 并利用该公式得到了有关“最大模”和“最大项”的相关估计.在此基础上, 文献[2-6]通过定义(R)级以及引入型函数定义的精确级研究了在半平面上解析的Laplace-Stieltjes增长性, 文[2]还通过定义指数级研究了在半平面上解析的具有相同零级的Laplace-Stieltjes变换的增长性, 得到了一些较好的结果.本文将通过引入对数级, 进一步精确地研究在右半平面上具有零级的Laplace-Stieltjes变换的增长性, 并得到关于这类变换的增长性的系数特征.考虑Laplace-Stieltjes变换所定义的函数

$\begin{aligned} F(s) = \int_0^{+\infty} {{{\rm{e}}^{ - sx}}{\rm{d}}\alpha (x)} \ \ (s=\sigma+it, \sigma, t \in \mathbb{R}), \end{aligned}$

(1)

$\alpha (x)$是对于$x \ge 0$有定义的实数或复数值函数, 而且它在任何闭区间$[0, X]$ $(0 < X < + \infty )$上是囿变的.

作序列

$\begin{aligned} 0 = {\lambda _0} < {\lambda _1} < {\lambda _2} < \cdots < {\lambda _n} \uparrow + \infty, \end{aligned}$

(2)

并且满足

$\begin{aligned} \overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{n}}{{{\lambda _n}}} =D<+ \infty, \overline {\mathop {\lim }\limits_{n \to \infty } } ({\lambda _{n + 1}} - {\lambda _n}) = h < + \infty. \end{aligned}$

(3)

令$A_n^* = \mathop {\sup }\limits_{{\lambda _n} < x \le {\lambda _{n + 1}}, t \in \mathbb{R}} \left| {\displaystyle\int_{{\lambda _n}}^x {{{\rm{e}}^{ - {\rm{i}}ty}}{\rm{d}}\alpha (y)} } \right|$, 由文[1]知, 当序列(2) 满足

$\begin{aligned} \overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{\log A_n^*}}{{{\lambda _n}}} = 0 \end{aligned}$

(4)

时, 变换(1) 的一致收敛横坐标为$\sigma _u^F = 0$, 此时变换(1) 定义的函数$F(s)$为右半平面上的解析函数.对$\sigma > 0$时, 记

$\begin{align} &{{M}_{u}}(\sigma ,F)=\underset{0<x<+\infty ,t\in \mathbb{R}}{\mathop{\sup }}\,\left| \int_{0}^{x}{{{\text{e}}^{-(\sigma +\text{i}t)y}}\text{d}\alpha (y)} \right|,\mu (\sigma ,F)=\underset{1\le n<+\infty }{\mathop{\max }}\,A_{n}^{*}{{\text{e}}^{-{{\lambda }_{n}}\sigma }}, \\ &N(\sigma )=\max \{n:\mu (\sigma ,F)=A_{n}^{*}{{\text{e}}^{-{{\lambda }_{n}}\sigma }}\}. \\ \end{align}$

定义 1.1 如果变换(1) 满足(2)--(4) 式, 定义变换(1) 在右半平面的增长级为

$\rho = \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{ - \log \sigma }}, $

当$\rho = 0$时, 称$F(s)$是右半平面上的零级L-S变换.

定义 1.2 设$F(s)$是右半平面上的零级L-S变换, 且存在$\varepsilon > 0$, 使得

$\begin{aligned} \frac{{\mu (\sigma, F)}}{{{\sigma ^{ - \varepsilon }}}} \to \infty (\sigma \to 0), \end{aligned}$

(5)

定义$F(s)$的对数级${\rho ^*}$和下对数级${\lambda ^*}$为

${\rho ^*} = \overline {\mathop {\lim }\limits_{\sigma \to 0 + } } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}}, {\lambda ^*} = \mathop {\underline {\lim } }\limits_{\sigma \to 0 + } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}}.$

引理 2.1^[2] 设变换(1) 满足(2)--(4) 式, 则对任意给定$\varepsilon \in (0, 1)$和充分接近0的$\sigma (\sigma > 0)$, 有

$\frac{1}{3}\mu (\sigma, F) \le {M_u}(\sigma, F) \le K(\varepsilon )\mu ((1-\varepsilon )\sigma, F)\frac{1}{\sigma },$

其中$K(\varepsilon )$是只与$\varepsilon$有关的常数.

注 1 由引理2.1及条件(5), 我们可以得到$1 \le {\lambda ^*}, {\rho ^*} \le + \infty $.

引理 2.2^[2] 设变换(1) 满足(2)--(4) 式, 则存在$\delta > 0$, 当$0 < \sigma < \delta$时,

$\log \mu (\sigma, F) = A - \int_\delta ^\sigma {{\lambda _{n(x)}}dx}, $

其中$A$为常数.

由文[7]中定理3.2的证明, 有

引理 2.3关天$x$的函数$g(x) = {(\log x)^T} + rx$的最大值为${\left[{\log \frac{T}{{-r}}} \right]^T} - T$, 其中$r < 0, x \ge 0$.

定理 3.1 设变换(1) 满足(2)--(5) 式, 且有对数级${\rho ^*}$, 则$\max \{ 1, \overline {\mathop {\lim }\limits_{n \to + \infty } } \frac{{{{\log }^ + }{{\log }^ + }A_n^*}}{{\log \log {\lambda _n}}}\} = {\rho ^*}.$

证 设$\max \{ 1, \overline {\mathop {\lim }\limits_{n \to + \infty } } \frac{{{{\log }^ + }{{\log }^ + }A_n^*}}{{\log \log {\lambda _n}}}\} = u$, 则$1 \le u \le \infty $.

先证明$u \le {\rho ^*}$, 不妨假设${\rho ^*} < + \infty $, 则$\forall \varepsilon > 0$, $\exists {\sigma _0} = {\sigma _0}(\varepsilon ) > 0$, 当$0 < \sigma < {\sigma _0}$时,

$\log {M_u}(\sigma, F) < {(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }}.$

由引理2.1, 则有$\log \mu (\sigma, F) < \log 3 + {(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }}, $所以$\log A_n^* < \log 3 + {(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }} + {\lambda _n}\sigma.$取$n$使得$\frac{1}{\sigma } = \frac{{{\lambda _n}}}{{{\rho ^*} + \varepsilon }}$, 则

$\log \log A_n^* < O(1) + ({\rho ^*} + \varepsilon )\log \log (\frac{{{\lambda _n}}}{{{\rho ^*} + \varepsilon }}) + {\rho ^*} + \varepsilon .$

所以由$\varepsilon $的任意性, 则有$\overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{{{\log }^ + }{{\log }^ + }A_n^*}}{{\log \log {\lambda _n}}} \le {\rho ^*}.$而当${\rho ^*} = + \infty $时上式显然成立, 故$u \le {\rho ^*}$.

下证$u \ge {\rho ^*}$, 不妨假设$u < + \infty $, 则对$\forall \varepsilon > 0$, $\exists {n_0} = {n_0}(\varepsilon ) > 0$, 当$n > {n_0}$时,

$\begin{aligned} \log A_n^* < {(\log {\lambda _n})^{u + \varepsilon }}. \end{aligned}$

(6)

另一方面,

$\int_0^x {{{\rm{e}}^{ - (\sigma + {\rm{i}}t)y}}{\rm{d}}\alpha (y)} = \sum\limits_{k = 1}^{n - 1} {\int_{{\lambda _k}}^{{\lambda _{k + 1}}} {{{\rm{e}}^{ - (\sigma + {\rm{i}}t)y}}{\rm{d}}\alpha (y)} + } \int_{{\lambda _n}}^x {{{\rm{e}}^{ - (\sigma + {\rm{i}}t)y}}{\rm{d}}\alpha (y)} \ \ ({\lambda _n} < x \le {\lambda _{n + 1}}), $

令

${I_k}(x, {\rm{i}}t) = \int_{{\lambda _k}}^x {{{\rm{e}}^{ - {\rm{i}}ty}}{\rm{d}}\alpha (y)} \ \ ({\lambda _n} < x \le {\lambda _{n + 1}}), $

则当${\lambda _k} \le x \le {\lambda _{k + 1}}, - \infty < t < + \infty $时

$\left| {{I_k}(x, {\rm{i}}t)} \right| \le A_k^* \le \mu (\sigma, F){{\rm{e}}^{{\lambda _k}\sigma }}.$

又由于

$\begin{array}{*{35}{l}} \int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{\text{e}}^{-\sigma y}}{{\text{d}}_{y}}{{I}_{k}}(y,\text{i}t)}&=\left. [{{\text{e}}^{-\sigma y}}{{I}_{k}}(y,\text{i}t)] \right|_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}-\int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{I}_{k}}(y,\text{i}t)\text{d}{{\text{e}}^{-\sigma y}}} \\ {}&={{\text{e}}^{-\sigma {{\lambda }_{k+1}}}}{{I}_{k}}({{\lambda }_{k+1}},\text{i}t)-{{\text{e}}^{-\sigma {{\lambda }_{k}}}}{{I}_{k}}({{\lambda }_{k}},\text{i}t)+\sigma \int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{I}_{k}}(y,\text{i}t){{\text{e}}^{-\sigma y}}\text{dy}} \\ {}&={{\text{e}}^{-\sigma {{\lambda }_{k+1}}}}{{I}_{k}}({{\lambda }_{k+1}},\text{i}t)+\sigma \int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{I}_{k}}(y,\text{i}t){{\text{e}}^{-\sigma y}}\text{dy}}, \\ \end{array}$

同理有

$\int_{{\lambda _n}}^x {{{\rm{e}}^{-\sigma y}}{{\rm{d}}_y}{I_n}(y, {\rm{i}}t)}= {{\rm{e}}^{-\sigma x}}{I_n}(x, {\rm{i}}t) + \sigma \int_{{\lambda _n}}^x {{I_n}(y, {\rm{i}}t){{\rm{e}}^{-\sigma y}}{\rm{dy}}}.$

于是当${\lambda _n} \le x \le {\lambda _{n + 1}}, \sigma > 0$时

$\begin{array}{*{35}{l}} \int_{0}^{x}{{{\text{e}}^{\text{-}(\sigma \text{+it})\text{y}}}\text{d}\alpha (\text{y})}&=&\sum\limits_{k=1}^{n-1}{\int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{\text{e}}^{\text{-}\sigma \text{y}}}{{\text{d}}_{\text{y}}}{{I}_{k}}(y,\text{it})}+}\int_{{{\lambda }_{n}}}^{x}{{{\text{e}}^{\text{-}\sigma \text{y}}}{{\text{d}}_{\text{y}}}{{I}_{n}}(y,\text{it})} \\ {}&=&\sum\limits_{k=1}^{n-1}{[{{\text{e}}^{\text{-}{{\lambda }_{\text{k+1}}}\sigma }}{{I}_{k}}({{\lambda }_{k+1}},\text{it})+\sigma \int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{\text{e}}^{\text{-}\sigma \text{y}}}{{I}_{k}}(y,\text{it})\text{dy}]}} \\ {}&{}&+{{\text{e}}^{-x\sigma }}{{I}_{n}}(x,\text{i}t)+\sigma \int_{{{\lambda }_{n}}}^{x}{{{\text{e}}^{-\sigma y}}{{I}_{n}}(y,\text{i}t)}\text{d}y, \\ \end{array}$

从而

$\begin{array}{*{35}{l}} \left| \int_{0}^{x}{{{\text{e}}^{\text{-}(\sigma \text{+it})\text{y}}}\text{d}\alpha (\text{y})} \right|&\le &\sum\limits_{k=1}^{n-1}{[{{\text{e}}^{\text{-}{{\lambda }_{\text{k+1}}}\sigma }}\left| {{I}_{k}}({{\lambda }_{k+1}},\text{it}) \right|+\sigma \int_{{{\lambda }_{k}}}^{{{\lambda }_{k+1}}}{{{\text{e}}^{\text{-}\sigma \text{y}}}\left| {{I}_{k}}(y,\text{it}) \right|\text{dy}]}} \\ {}&{}&+{{\text{e}}^{-x\sigma }}\left| {{I}_{n}}(x,\text{i}t) \right|+\sigma \int_{{{\lambda }_{n}}}^{x}{{{\text{e}}^{-\sigma y}}\left| {{I}_{n}}(y,\text{i}t) \right|}\text{d}y \\ {}&\le &\sum\limits_{k=1}^{n-1}{[A_{k}^{*}{{\text{e}}^{\text{-}{{\lambda }_{\text{k+1}}}\sigma }}+A_{k}^{*}({{\text{e}}^{\text{-}\sigma {{\lambda }_{\text{k}}}}}-{{\text{e}}^{\text{-}\sigma {{\lambda }_{\text{k+1}}}}})]}+A_{n}^{*}{{\text{e}}^{\text{-}\sigma {{\lambda }_{\text{n}}}}} \\ {}&\le &\sum\limits_{n=1}^{\infty }{A_{n}^{*}{{\text{e}}^{\text{-}{{\lambda }_{\text{n}}}\sigma }}}. \\ \end{array}$

所以由(6) 式

${M_u}(\sigma, F) \le\sum\limits_{n = 1}^{{n_0}} {A_n^*{{\rm{e}}^{ - {\lambda _n}\sigma }}} + \sum\limits_{n = {n_0} + 1}^{ + \infty } {A_n^*{{\rm{e}}^{ - {\lambda _n}\sigma }}}\le P({n_0}) + \sum\limits_{n = {n_0} + 1}^\infty {\exp \{ } {(\log {\lambda _n})^{u + \varepsilon }} - {\lambda _n}\sigma \},$

(7)

其中$P({n_0})$是与${n_0}$有关的常数.取$N$满足$N = (D + \varepsilon )\frac{2}{\sigma }{(\log \frac{2}{\sigma })^{u + \varepsilon }}$, 由(7) 式和引理2.3有

${M_u}(\sigma, F) \le P({n_0}) + \sum\limits_{n = {n_0} + 1}^N {\exp \{ {{(\log {\lambda _n})}^{u + \varepsilon }} - {\lambda _n}\sigma \} } + \sum\limits_{n = N + 1}^\infty {\exp \{ {{(\log {\lambda _n})}^{u + \varepsilon }} - {\lambda _n}\sigma \} } \\ \le P({n_0}) + N(\exp \{ {(\log \frac{{u + \varepsilon }}{\sigma })^{u + \varepsilon }} - (u + \varepsilon )\} ) + \sum\limits_{n = N + 1}^\infty {\exp \{ {{(\log {\lambda _n})}^{u + \varepsilon }} - {\lambda _n}\sigma \} }.$

对于每个$\sigma > 0$, 定义函数$n(\sigma )$满足${\lambda _{n(\sigma )}} < \frac{2}{\sigma } < {\lambda _{n(\sigma ) + 1}}$, 则对充分接近0的$\sigma (\sigma > 0)$, 由(3) 式中第一个等式, 当$n > N$时,

${\lambda _n} > \frac{n}{{D + \varepsilon }} > \frac{N}{{D + \varepsilon }} = \frac{2}{\sigma }{(\log \frac{2}{\sigma })^{u + \varepsilon }} > \frac{2}{\sigma }{(\log {\lambda _n})^{u + \varepsilon }}.$

所以

$\quad\sum\limits_{n = N + 1}^\infty {\exp \{ {{(\log {\lambda _n})}^{u + \varepsilon }} - {\lambda _n}\sigma \} } \le \sum\limits_{n = N + 1}^\infty {\exp \{ - \frac{{{\lambda _n}\sigma }}{2}\} } \\ \le \sum\limits_{n = N + 1}^\infty {\exp \{ - \frac{{n\sigma }}{{2(D + \varepsilon )}}\} } = \frac{{\exp \{ \frac{{ - \sigma (N + 1)}}{{2(D + \varepsilon )}}\} }}{{1 - \exp \{ \frac{{ - \sigma }}{{2(D + \varepsilon )}}\} }} \le \frac{{\exp \{ \frac{{ - \sigma N}}{{2(D + \varepsilon )}}\} }}{{1 - \exp \{ \frac{{ - \sigma }}{{2(D + \varepsilon )}}\} }}.$

由于

$\mathop {\lim }\limits_{\sigma \to 0} \frac{{\exp \{ \frac{{-\sigma N}}{{2(D + \varepsilon )}}\} }}{{1-\exp \{ \frac{{-\sigma }}{{2(D + \varepsilon )}}\} }} = \mathop {\lim }\limits_{\sigma \to 0} \frac{{\exp \{ - {{(\log \frac{2}{\sigma })}^{u + \varepsilon }}\} }}{{\frac{\sigma }{{2(D + \varepsilon )}}(1 + o(\sigma ))}} = 0,$

所以

$\sum\limits_{n = N + 1}^\infty {\exp \{ {{(\log {\lambda _n})}^{u + \varepsilon }} - {\lambda _n}\sigma \} } = o(1)\ \ (\sigma \to 0) .$

所以对充分接近0的$\sigma (\sigma > 0)$,

${M_u}(\sigma, F) \le P({n_0}) + N(\exp \{ {(\log \frac{{u + \varepsilon }}{\sigma })^{u + \varepsilon }}-(u + \varepsilon )\} ) + o(1),$

则

$\log \log {M_u}(\sigma, F) \le (1-o(1))(u + \varepsilon )\log \log \frac{{u + \varepsilon }}{\sigma } + o(1),$

由$\varepsilon$的任意性, 有

$\overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{\log \log {M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}} \le u.$

由于$u = + \infty $时上式是显然成立的, 所以${\rho ^*} \le u$.

定理 3.2 设变换(1) 满足(2)--(5) 式, 且有对数级${\rho ^*}$, 则

$\overline {\mathop {\lim }\limits_{\sigma \to 0} }\frac{{{{\log }^ + }{{\log }^ + }\mu (\sigma, F)}}{{\log \log \frac{1}{\sigma }}} = \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}} = {\rho ^*}.$

证 作Dirichlet级数

$f(s) = \sum\limits_{n = 1}^{ + \infty } {A_n^*{{\rm{e}}^{ - {\lambda _n}s}}},$

(8)

则级数(8) 所确定的函数是右半平面上的解析函数, 由文[8]中定理1.1和定理1.2有

$\max \{ 1, \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }A_n^*}}{{\log \log {\lambda _n}}}\} = \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }\mu (\sigma, F)}}{{\log \log \frac{1}{\sigma }}},$

再由定理3.1, 可知定理成立.

定理 3.3 设变换(1) 满足(2)--(5) 式, 且有对数级${\rho ^*}(1 < {\rho ^*} < + \infty )$, 则\[{\rho ^*}-1 \le \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{\log ({\lambda _{N(\sigma)}}\sigma)}}{{\log \log \frac{1}{\sigma }}}\} \le {\rho ^*}.\]

证 设$\overline {\mathop {\lim }\limits_{\sigma \to 0}} \frac{{\log ({\lambda _{N(\sigma )}}\sigma )}}{{\log \log \frac{1}{\sigma }}}\} = \theta $, 则$\forall \varepsilon > 0, \exists {\sigma _1} = {\sigma _1}(\varepsilon ) > 0$, 当$0 < \sigma < {\sigma _1}$时,

$\begin{aligned} {\lambda _{N(\sigma )}}\sigma < {(\log \frac{1}{\sigma })^{\theta + \varepsilon }}, \end{aligned}$

(9)

由引理2.2,

$\log \mu (\sigma, F) = A-\int_{{\sigma _0}}^\sigma {{\lambda _{N(x)}}{\rm{d}}x}.$

令${\sigma _0} = \min \{ \delta, {\sigma _1}\} $, 则当$0 < \sigma < {\sigma _0}$时, 由(9) 式有

$\begin{array}{*{20}{l}} {\log \mu (\sigma ,F)}&{ < A - \int_{{\sigma _0}}^\sigma {\frac{{{{(\log \frac{1}{x})}^{\theta + \varepsilon }}}}{x}{\rm{d}}x} }\\ {}&{ \le A + \frac{1}{{\theta + \varepsilon + 1}}[{{(\log \frac{1}{\sigma })}^{\theta + \varepsilon + 1}} - {{(\log \frac{1}{{{\sigma _0}}})}^{\theta + \varepsilon + 1}}]}\\ {}&{ \le \frac{1}{{\theta + \varepsilon + 1}}{{(\log \frac{1}{\sigma })}^{\theta + \varepsilon + 1}} + O(1).} \end{array}$

由$\varepsilon$的任意性, 有

$\overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }\mu (\sigma, F)}}{{\log \log \frac{1}{\sigma }}} \le \theta + 1,$

再由定理3.2, 则有${\rho ^*} \le \theta + 1$.

另一方面, 由${\rho ^*} = \overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}}$, 则$\forall \varepsilon > 0, \exists {\sigma _1} = {\sigma _1}(\varepsilon ) > 0$, 当$0 < \sigma < {\sigma _1}$时,

$\log {M_u}(\sigma, F) < {(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }}.$

由引理1.1, 则有

$\begin{aligned} \log \mu (\sigma, F) < \log 3 + {(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }}, \end{aligned}$

(10)

由(3) 式, 对上述的$\varepsilon $和$\sigma $有

$\sigma N(\sigma ) < (D + \varepsilon ){\lambda _{N(\sigma )}}\sigma \le-2(D + \varepsilon )\int_\sigma ^{\frac{\sigma }{2}} {{\lambda _{N(x)}}{\rm{d}}x} \le 2(D + \varepsilon )\log \mu (\sigma, F) + O(1).$

再由(10) 式, 则有

$\sigma N(\sigma ) \le 2(D + \varepsilon ){(\log \frac{1}{\sigma })^{{\rho ^*} + \varepsilon }} + O(1),$

两边取对数再除以$\log \log \frac{1}{\sigma }$, 然后取上极限, 则有

$\overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{\log ({\lambda _{N(\sigma )}}\sigma )}}{{\log \log \frac{1}{\sigma }}} \le {\rho ^*} + \varepsilon,$

由$\varepsilon$的任意性, 有

$\overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{\log ({\lambda _{N(\sigma )}}\sigma )}}{{\log \log \frac{1}{\sigma }}} \le {\rho ^*}.$

综合前半部分证明, 可知定理成立.

注 2 若$\overline {\mathop {\lim }\limits_{\sigma \to 0} } \frac{{\log ({\lambda _{N(\sigma )}}\sigma )}}{{\log \log \frac{1}{\sigma }}} = 0$, 由定理3.2和注1可知此时${\rho ^*} = 1$.

定理 3.4 设变换(1) 满足(2)--(5) 式, 且有下对数级${\lambda ^*}(1 \le {\lambda ^*} \le \infty )$, $\{ {n_k}\} _1^\infty $是一列递增正整数序列, 则

$\mathop {\underline {\lim } }\limits_{k \to \infty } \frac{{{{\log }^ + }{{\log }^ + }A_{{n_k}}^*}}{{\log \log {\lambda _{{n_{k + 1}}}}}} \le {\lambda ^*}.$

证 设$\mathop {\underline {\lim } }\limits_{k \to \infty } \frac{{{{\log }^ + }{{\log }^ + }A_{{n_k}}^*}}{{\log \log {\lambda _{{n_{k + 1}}}}}} = \theta (0 \le \theta < \infty )$, 则对任意给定的$\varepsilon > 0$, 存在$N > 0$, 当$k > N$时,

${\log ^ + }A_{{n_k}}^* \ge {(\log {\lambda _{{n_{k + 1}}}})^{\theta-\varepsilon }}.$

选取${\sigma _k} = \frac{{\theta - \varepsilon }}{{{\lambda _{{n_k}}}}}, k = 1, 2, \cdots $, 当$k > N$且${\sigma _{k + 1}} \le \sigma \le {\sigma _k}$时,

$\begin{array}{*{20}{l}} {{{\log }^ + }{M_u}(\sigma ,F) \ge {{\log }^ + }\mu (\sigma ,F)}&{ \ge {{(\log {\lambda _{{n_{k + 1}}}})}^{\theta - \varepsilon }} - {\lambda _{{n_k}}}\sigma }\\ {}&\begin{array}{l} \ge {(\log \frac{1}{{{\sigma _{k + 1}}}})^{\theta - \varepsilon }} + O(1) \ge {(\log \frac{1}{\sigma })^{\theta - \varepsilon }} + O(1),\\ \end{array} \end{array}$

所以

${\lambda ^*} = \mathop {\underline {\lim } }\limits_{\sigma \to 0 + } \frac{{{{\log }^ + }{{\log }^ + }{M_u}(\sigma, F)}}{{\log \log \frac{1}{\sigma }}} \ge \theta.$

若$\theta = \infty $, 显然${\lambda ^*} = \infty $, 所以定理得证.